Entropy in Quantum Statistical Mechanics

The von Neumann formula

$$S(\rho) = -k_{\mathrm B} \operatorname{Tr} \left( \rho\ln\rho \right)$$

appears in quantum information, equilibrium statistical mechanics, entanglement theory, and nonequilibrium thermodynamics. The formula is the same, but its physical interpretation is not fixed by the formula alone. One must identify the state, the system boundary, the subsystem partition, the constraints, and any information that has been discarded.

For the Gibbs state of a temperature-independent Hamiltonian, $S(\rho)$ agrees with the entropy derived from its canonical free energy. For the reduced state of part of a globally pure system, it is entanglement entropy. For a generic mixed subsystem, it measures local mixedness and can contain thermal uncertainty, classical correlation, entanglement, and environmental noise. For a deliberately coarse-grained state, it depends on which distinctions were erased.

The examples below compare the full state with a reduced, dephased, or coarse-grained state. For spatial scaling laws and numerical methods, continue to many-body entanglement entropy.

Helpful background. Entropy of a density operator supplies the eigenvalue formula; thermal density operators supply the equilibrium states.

Entropy of the state spectrum

It is useful to separate the dimensionless state entropy

$$\mathsf S(\rho) = - \operatorname{Tr} \left( \rho\ln\rho \right)$$

from the physical entropy

$$S(\rho) = k_{\mathrm B}\mathsf S(\rho).$$

With the natural logarithm, $\mathsf S$ is measured in nats. If $\mathsf S_2$ instead uses $\log_2$, it is measured in bits and $S=k_{\mathrm B}\ln2\,\mathsf S_2$. Multiplying a bit count by $k_{\mathrm B}$ alone misses the factor $\ln2$.

If

$$\rho = \sum_j p_j |j\rangle\langle j|$$

is the spectral decomposition, then

$$\mathsf S(\rho) = - \sum_j p_j\ln p_j.$$

Thus the von Neumann entropy is the Shannon entropy of the eigenvalue distribution. It is basis independent:

$$\mathsf S(U\rho U^\dagger) = \mathsf S(\rho).$$

The entropy is a nonlinear functional of the state, not the expectation value of a fixed observable. There is no state-independent Hermitian operator $\hat S$ satisfying $S(\rho)=\operatorname{Tr}(\rho\hat S)$ for all $\rho$.

When State Entropy Is Thermodynamic Entropy

The identification

$$S_{\mathrm{th}} = S(\rho_{\mathrm{eq}})$$

is justified when $\rho_{\mathrm{eq}}$ is the equilibrium state associated with the macroscopic constraints and when the resulting entropy obeys the thermodynamic relations appropriate to those controls.

For a regular equilibrium family, this includes relations such as

$$\frac{1}{T} = \left( \frac{\partial S}{\partial U} \right)_{V,N},$$ $$S = - \left( \frac{\partial F}{\partial T} \right)_{V,N},$$

and

$$S = - \left( \frac{\partial\Omega_{\mathrm G}} {\partial T} \right)_{V,\mu}.$$

The last two derivative formulas assume the microscopic Hamiltonian has no unaccounted explicit temperature dependence. See thermodynamic potentials for these derivative conventions. In particular, the von Neumann entropy of a strongly coupled subsystem generally differs from the thermodynamic entropy derived from its mean-force free energy: its effective Hamiltonian depends on temperature.

The equality between state entropy and thermodynamic entropy is not a claim that every density operator is an equilibrium macrostate. A qubit prepared by a noisy device has a von Neumann entropy, but no temperature, pressure, or entropy density follows without further physical structure.

Canonical thermal entropy

For $\rho_{N,\beta}=e^{-\beta H_N}/Z_N$, inserting $\ln\rho=-\beta H_N-\ln Z_N$ into the entropy definition gives

$$S_{\mathrm c}=k_{\mathrm B}(\ln Z_N+\beta U)=\frac{U-F}{T}.$$

If $H_N$ is temperature independent and the necessary moments are finite, this equals $-(\partial F/\partial T)_{V,N}$. The exercise below verifies the derivative explicitly. At positive temperature, $\partial_T S=C_{V,N}/T\geq0$; the canonical ensemble derives the heat-capacity identity.

Entropy across particle-number sectors

For the grand-canonical state,

$$S_{\mathrm{gc}}=k_{\mathrm B}[\ln\Xi+\beta(U-\mu\overline N)] =\frac{U-\mu\overline N-\Omega_{\mathrm G}}{T}.$$

There is a useful second expression. Because $\rho=\bigoplus_N P(N)\rho_N$ has eigenvalues $P(N)p_{j|N}$,

$$S(\rho)=-k_{\mathrm B}\sum_N P(N)\ln P(N) +\sum_N P(N)S(\rho_N).$$

The first term measures uncertainty about the number sector; the second averages the uncertainty within each sector. Hence an exact grand-canonical entropy is not obtained by substituting the mean number into the canonical entropy. See grand-canonical ensembles for $P(N)$ and the conditions for local ensemble equivalence.

Microcanonical Entropy

Let $P_{E,\Delta}$ project onto an energy shell and let

$$W(E,\Delta) = \operatorname{Tr} P_{E,\Delta}$$

be its dimension. The microcanonical state is

$$\rho_{\mathrm{mc}} = \frac{P_{E,\Delta}}{W(E,\Delta)}.$$

Its $W$ nonzero eigenvalues all equal $1/W$, so

$$S(\rho_{\mathrm{mc}}) = k_{\mathrm B} \ln W(E,\Delta).$$

This is the shell-count form of Boltzmann entropy. It is exactly equal to the von Neumann entropy of the uniform shell state.

The equality does not assign entropy $k_{\mathrm B}\ln W$ to every pure state inside the shell. A pure energy-shell vector obeys

$$S \left( |\psi\rangle\langle\psi| \right) = 0,$$

even when local observables in $|\psi\rangle$ resemble microcanonical predictions. Typicality or the eigenstate thermalization hypothesis can explain agreement for restricted observables; neither changes the global pure-state entropy.

Shell-count, density-of-states, and cumulative-count entropies can differ at finite size. The microcanonical ensemble explains those conventions and the energy-window width.

Free Energy and Relative Entropy

For a canonical reference state $\rho_\beta$, define

$$\mathcal F_T(\rho) = \operatorname{Tr}(\rho\hat H) - T S(\rho).$$

Then

$$\mathcal F_T(\rho) - F_{\mathrm{eq}} = k_{\mathrm B}T D(\rho\Vert\rho_\beta),$$

where

$$D(\rho\Vert\sigma) = \operatorname{Tr} \left[ \rho \left( \ln\rho- \ln\sigma \right) \right].$$

At positive $T$, nonnegativity of relative entropy gives

$$\mathcal F_T(\rho) \geq F_{\mathrm{eq}}.$$

This identity connects an information-theoretic distinguishability to a thermodynamic free-energy excess. It does not by itself specify how much work a restricted experimental protocol can extract. For the mathematical divergence see relative entropy; for work protocols see quantum thermodynamics. At a normalizable negative temperature, the same identity reverses the inequality.

State Entropy Versus Measurement Entropy

For a projective measurement with outcomes $a$,

$$p_a = \operatorname{Tr}(\rho P_a),$$

the outcome entropy is

$$H_{\mathcal M}(\rho) = - \sum_a p_a\ln p_a.$$

This is the Shannon entropy of one chosen measurement distribution. It depends on the measurement, whereas $\mathsf S(\rho)$ depends only on the state spectrum.

For a rank-one orthonormal projective measurement,

$$H_{\mathcal M}(\rho) \geq \mathsf S(\rho),$$

with equality when the measurement diagonalizes $\rho$, up to degeneracies. A pure state has zero von Neumann entropy but can have a highly uncertain measurement outcome in an incompatible basis.

The rank-one condition matters. A degenerate measurement can have only one certain outcome while the state remains mixed inside its eigenspace; its outcome entropy can then be smaller than the state entropy.

Thus “uncertainty” must be qualified:

• state entropy describes spectral mixedness;
• measurement entropy describes uncertainty in a specified outcome distribution;
• thermodynamic entropy describes an equilibrium state relative to macroscopic controls;
• entanglement entropy describes a reduced state relative to a subsystem split.

See Shannon entropy for the classical outcome measure.

Additivity, Correlations, and Extensivity

For an exact product state,

$$\rho_{AB} = \rho_A\otimes\rho_B,$$

von Neumann entropy is additive:

$$S(\rho_A\otimes\rho_B) = S(\rho_A) + S(\rho_B).$$

For a correlated state, subadditivity gives

$$S(\rho_{AB}) \leq S(\rho_A) + S(\rho_B).$$

The difference is the mutual information,

$$I(A:B) = S(\rho_A) + S(\rho_B) - S(\rho_{AB}).$$

Equivalently,

$$I(A:B) = k_{\mathrm B} D \left( \rho_{AB} \Vert \rho_A\otimes\rho_B \right) \geq 0.$$

Thermodynamic extensivity is a scaling property of a family of systems, distinct from the exact additivity identity for a product state. Interacting equilibrium systems generally have correlations across boundaries. For short-range interactions in a regular thermodynamic limit,

$$S_L = s_{\mathrm{th}}V_L + o(V_L),$$

so the leading term is extensive while boundaries and correlations contribute subextensive corrections. Long-range interactions, interfaces, constraints, and criticality can require additional care.

Thermal Entropy Versus Entanglement Entropy

For a pure bipartite state $|\Psi\rangle_{AB}$,

$$S(\rho_{AB})=0,$$

but the reduced states

$$\rho_A = \operatorname{Tr}_B |\Psi\rangle\langle\Psi|$$

and $\rho_B$ can be mixed. Their common entropy,

$$S(\rho_A) = S(\rho_B),$$

is the bipartite entanglement entropy.

For a mixed thermal state $\rho_{AB,\beta}$, the subsystem entropy $S(\rho_A)$ is not a general entanglement measure. It can contain:

• local thermal mixing;
• entanglement across the cut;
• classical correlations;
• correlations with an external environment;
• boundary and finite-size contributions.

In a homogeneous thermal phase, a sufficiently large region can have

$$S(\rho_A) = s_{\mathrm{th}}|A| + \text{boundary corrections}.$$

The volume term is thermal entropy, not automatically bipartite entanglement.

A globally pure, highly excited many-body eigenstate presents a different case:

$$S \left( |E\rangle\langle E| \right) = 0,$$

while a subregion can have volume-law entanglement entropy close to the thermal entropy of that subregion. This is a local statement about $\rho_A$, not a conversion of the global pure state into a thermal mixture.

For the pure bipartite measure see entanglement entropy; for mixed-state marginals see subsystem entropy. Spatial area and volume laws are developed in many-body entanglement.

Same Local Entropy, Different Global Physics

Consider three two-qubit states.

First, the Bell state

$$|\Phi^+\rangle = \frac{ |00\rangle+|11\rangle }{\sqrt2}$$

is globally pure. Second, the classically correlated mixture is

$$\rho_{\mathrm{cc}} = \frac12 |00\rangle\langle00| + \frac12 |11\rangle\langle11|.$$

Third, the uncorrelated maximally mixed state is

$$\rho_{\mathrm{prod}} = \frac{\mathbb I_A}{2} \otimes \frac{\mathbb I_B}{2}.$$

All three have the same one-qubit reduced state:

$$\rho_A = \rho_B = \frac{\mathbb I}{2},$$

and therefore

$$S_A = S_B = k_{\mathrm B}\ln2.$$

Their global entropies and mutual informations differ:

State	$S_{AB}$	$I(A:B)$	Correlation type
Bell state	$0$	$2k_{\mathrm B}\ln2$	pure-state entanglement
classically correlated mixture	$k_{\mathrm B}\ln2$	$k_{\mathrm B}\ln2$	classical correlation
product maximally mixed state	$2k_{\mathrm B}\ln2$	$0$	no $A$–$B$ correlation

The local entropy alone cannot reveal which global state produced it. The state boundary and correlation measure are indispensable.

Fine-Grained Entropy Under Unitary Evolution

For a closed system,

$$\rho(t) = U(t)\rho(0)U^\dagger(t).$$

Unitary evolution preserves the eigenvalues of $\rho$, so

$$S(\rho(t)) = S(\rho(0)).$$

This exact conservation does not contradict thermodynamic entropy increase. The two statements refer to different descriptions or boundaries.

Entropy can change when one:

• traces out an environment or complement;
• discards phase information by dephasing;
• replaces the exact state by a coarse-grained macrostate;
• restricts attention to a set of observables;
• uses an open-system dynamical map;
• takes a thermodynamic or long-time limit before reversing the dynamics.

None of these operations is the exact unitary evolution of the complete state. Moreover, subsystem von Neumann entropy need not increase monotonically, and not every open-system channel increases entropy. The second law requires an entropy balance that includes reservoirs, correlations, and the allowed operations. See entropy production for those dynamical inequalities.

Dephasing and Diagonal Entropy

Let

$$\hat H = \sum_E E P_E$$

be the spectral decomposition, including degenerate eigenspaces. Energy dephasing maps

$$\rho \longmapsto \mathcal D_H(\rho) = \sum_E P_E\rho P_E.$$

This removes coherence between distinct energies while retaining coherence inside degenerate energy subspaces. In finite dimension, $\mathcal D_H$ is a unital quantum channel, and

$$S \left( \mathcal D_H(\rho) \right) \geq S(\rho).$$

For a nondegenerate Hamiltonian, the entropy of the dephased state is the diagonal entropy,

$$S_{\mathrm d} = -k_{\mathrm B} \sum_n \rho_{nn} \ln\rho_{nn}.$$

A pure superposition can have $S(\rho)=0$ and $S_{\mathrm d}>0$. The increase records discarded phase coherence, not a change in the exact global state.

Diagonal entropy can approach thermodynamic entropy for broad classes of large nonintegrable systems and suitable preparations, but that connection has hypotheses. Integrability, degeneracies, conserved charges, and narrow or structured energy distributions can obstruct a simple identification.

Projective Coarse-Grained Entropy

Consider a projective coarse-graining $\mathcal C=\{P_\alpha\}$ with

$$\sum_\alpha P_\alpha = \mathbb I.$$

Define the macrostate probability and Hilbert-space volume by

$$p_\alpha = \operatorname{Tr} \left( P_\alpha\rho \right), \qquad V_\alpha = \operatorname{Tr} P_\alpha.$$

The single-coarse-graining observational entropy is

$$S_{\mathcal C}(\rho) = -k_{\mathrm B} \sum_\alpha p_\alpha \ln \left( \frac{p_\alpha}{V_\alpha} \right).$$

It is the von Neumann entropy of the block-uniform coarse state

$$\rho_{\mathcal C} = \sum_\alpha \frac{p_\alpha}{V_\alpha} P_\alpha.$$

Therefore, in finite dimension,

$$S(\rho) \leq S_{\mathcal C}(\rho) \leq k_{\mathrm B} \ln d,$$

where $d$ is the Hilbert-space dimension.

If the state is known only to occupy one macrospace $\alpha$, then $p_\alpha=1$ and

$$S_{\mathcal C} = k_{\mathrm B} \ln V_\alpha.$$

This recovers the Boltzmann form. A microcanonical shell is one physically important macrospace.

In fact the entropy increase has the exact expression

$$S_{\mathcal C}(\rho)-S(\rho) =k_{\mathrm B}D(\rho\Vert\rho_{\mathcal C}).$$

To see why, $\ln\rho_{\mathcal C}$ is constant on each occupied macrospace. Its expectation therefore depends only on the probabilities $p_\alpha$, which the coarse-graining preserves. This proves the lower bound and shows that equality holds exactly when the original state is already block uniform.

The coarse-graining is part of the definition. Position cells, energy windows, local densities, and experimentally resolved observables give different entropies. A coarse-grained entropy is trustworthy only when the retained distinctions match the physical question and resolution.

More elaborate observational entropies can use sequences of noncommuting coarse-grainings. Their ordering and interpretation require additional care; the single projective construction above is the clean benchmark.

Constraint-Based Coarse Graining

Another common coarse-grained entropy keeps only selected expectation values $q_a$ and maximizes over all compatible states:

$$S_{\mathrm{cg}} \left( \{q_a\} \right) = \sup_{\rho} \left\{ S(\rho) : \operatorname{Tr}(\rho\hat Q_a)=q_a \right\}.$$

The exact state entropy obeys

$$S(\rho) \leq S_{\mathrm{cg}}$$

whenever $\rho$ satisfies the retained constraints. Equality means that the exact state already is the maximum-entropy representative of those data.

This construction makes the dependence on retained information explicit. Adding constraints can only keep the maximum entropy unchanged or lower it, because the feasible set becomes smaller.

Projective observational entropy, diagonal entropy, maximum-entropy inference, and subsystem entropy are related ways of discarding distinctions, but they are not interchangeable definitions. Each specifies a different information-losing map or optimization.

Entropy Density and the Thermodynamic Limit

For a sequence of finite systems of volume $V_L$, a thermodynamic entropy density is

$$s = \lim_{L\to\infty} \frac{S_L}{V_L},$$

when the limit exists under specified boundary conditions and fixed intensive variables.

The density can remain finite even when the total entropy diverges with volume. In an infinite system, the total density operator may not be trace class and the total von Neumann entropy may not exist. Local entropies, entropy densities, relative entropies, and differences are often the better-defined objects.

Spatial entanglement entropy in a continuum can also diverge with the ultraviolet cutoff. That divergence is not the same as the extensive thermal divergence with system volume. The regulator, region geometry, and order of limits must be stated.

See the thermodynamic limit for the limiting procedure and boundary assumptions.

Zero Temperature and Residual Entropy

For a finite-dimensional system with a unique ground state,

$$\lim_{T\to0^+} \rho_\beta = |0\rangle\langle0|,$$

so

$$\lim_{T\to0^+} S(\rho_\beta) = 0.$$

If the exact ground space has degeneracy $g$, the finite-system Gibbs limit is the uniform ground-space mixture,

$$\rho_{0^+} = \frac{P_0}{g},$$

and

$$S(\rho_{0^+}) = k_{\mathrm B}\ln g.$$

An infinitesimal symmetry-breaking field, a superselection sector, or a different order of thermodynamic and zero-temperature limits can select a pure phase instead. Residual entropy statements must therefore specify degeneracy, preparation, sectors, and limit order.

Negative Temperature Does Not Mean Negative Entropy

For a bounded spectrum, a microcanonical entropy can decrease with energy near the upper spectral edge. Then

$$\frac{1}{T} = \left( \frac{\partial S}{\partial U} \right)_{V,N}$$

can be negative.

The entropy itself remains nonnegative for a finite-dimensional density operator:

$$0 \leq S(\rho) \leq k_{\mathrm B}\ln d.$$

Negative temperature refers to the slope of the entropy, not to a negative entropy value. Finite-system conclusions can depend on the chosen microcanonical entropy convention, as discussed on the Microcanonical Ensemble page.

Infinite-Dimensional Caveats

In infinite-dimensional Hilbert space, a normalized state can have infinite von Neumann entropy. A canonical partition function can converge while additional moments or entropy limits still require checking. For a Gibbs state, finiteness of both

$$Z = \operatorname{Tr} e^{-\beta\hat H}$$

and

$$U = \frac{ \operatorname{Tr} \left( \hat H e^{-\beta\hat H} \right) }{Z}$$

is sufficient for the standard expression

$$S = k_{\mathrm B} \left( \ln Z+\beta U \right)$$

to be finite.

In continuum quantum field theory, local algebras need not factor into ordinary tensor products and spatial entanglement entropy is regulator dependent. Finite-dimensional formulas remain useful guides, but they should not be transferred to the continuum without the required algebraic and ultraviolet qualifications.

Exercises

State entropy and measurement entropy

A qubit is in the pure state $|0\rangle$. Compute its von Neumann entropy and the Shannon entropy of measurements in the $z$ and $x$ bases.

Solution

The density operator is

$$\rho = |0\rangle\langle0|,$$

whose eigenvalues are $1$ and $0$. Therefore

$$S(\rho)=0.$$

In the $z$ basis, the outcomes have probabilities $(1,0)$, so

$$H_z=0.$$

In the $x$ basis,

$$|0\rangle = \frac{ |+\rangle+|-\rangle }{\sqrt2},$$

so the outcomes have probabilities $(1/2,1/2)$. Hence

$$H_x = -2 \left( \frac12\ln\frac12 \right) = \ln2.$$

The state entropy remains zero in both cases. Only the measurement entropy changes.

Canonical entropy identity

Starting from

$$\rho_\beta = \frac{e^{-\beta\hat H}}{Z},$$

derive

$$S = k_{\mathrm B} \left( \ln Z+\beta U \right)$$

and show that $S=-\partial_TF$ for a temperature-independent Hamiltonian.

Solution

Since

$$\ln\rho_\beta = -\beta\hat H- \ln Z,$$

we have

$$\begin{aligned} S &= -k_{\mathrm B} \operatorname{Tr} \left( \rho_\beta\ln\rho_\beta \right) \\ &= k_{\mathrm B} \left[ \beta \operatorname{Tr} \left( \rho_\beta\hat H \right) + \ln Z \right] \\ &= k_{\mathrm B} \left( \ln Z+\beta U \right). \end{aligned}$$

Now

$$F = -k_{\mathrm B}T\ln Z.$$

For temperature-independent $\hat H$,

$$\frac{\partial\ln Z}{\partial T} = \frac{U}{k_{\mathrm B}T^2}.$$

Therefore

$$\begin{aligned} - \frac{\partial F}{\partial T} &= k_{\mathrm B}\ln Z + k_{\mathrm B}T \frac{\partial\ln Z}{\partial T} \\ &= k_{\mathrm B}\ln Z + \frac{U}{T} \\ &= S. \end{aligned}$$

Resolution dependence of coarse entropy

Let a four-dimensional Hilbert space have basis $\{|1\rangle,|2\rangle,|3\rangle,|4\rangle\}$. The exact state is $|1\rangle$. A coarse-graining has two macrospaces,

$$P_A = |1\rangle\langle1| + |2\rangle\langle2|,$$

and

$$P_B = |3\rangle\langle3| + |4\rangle\langle4|.$$

Compute the observational entropy. Then refine $P_A$ into one-dimensional projectors and compute it again.

Solution

For the coarse partition,

$$p_A=1, \qquad p_B=0,$$

and

$$V_A=V_B=2.$$

Thus

$$S_{\mathcal C} = -k_{\mathrm B} \ln \left( \frac12 \right) = k_{\mathrm B}\ln2.$$

The observer knows only that the state lies in a two-dimensional macrospace.

After refining $P_A$ into $|1\rangle\langle1|$ and $|2\rangle\langle2|$, the occupied macrospace has volume one. Hence

$$S_{\mathcal C'}=0.$$

The exact state has zero von Neumann entropy throughout. The coarse entropy changes because the retained resolution changes.

Dephasing is not uniform mixing within a degenerate block

In a three-dimensional space let $H=\Delta\lvert3\rangle\langle3\rvert$ and $\lvert\psi\rangle=(\lvert1\rangle+\lvert2\rangle)/\sqrt2$. Compare the energy-measurement outcome entropy, the entropy after energy dephasing, and the projective coarse-grained entropy for $P_0=\lvert1\rangle\langle1\rvert+\lvert2\rangle\langle2\rvert$ and $P_\Delta=\lvert3\rangle\langle3\rvert$.

Solution

The energy is certainly zero, so the outcome entropy is zero. Energy dephasing leaves the pure state unchanged because both components lie in the same degenerate energy block; its state entropy is also zero. The block-uniform coarse state is instead $P_0/2$ and has entropy $k_{\mathrm B}\ln2$. Resolving an energy value, erasing coherences between different energies, and forgetting all distinctions within an energy block are three different operations.

References

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